Introduction to Econometrics, 3e (Stock)Chapter 2 Review of Probability2.1 Multiple Choice1) The probability of an outcomeA) is the number of times that the outcome occurs in the long run.B) equals M N, where M is the number of occurrences and N is the population size.C) is the proportion of times that the outcome occurs in the long run.D) equals the sample mean divided by the sample standard deviation.Answer:C2) The probability of an event A or B (Pr(A or B)) to occur equalsA) Pr(A) Pr(B).B) Pr(A) + Pr(B) if A and B are mutually exclusive.C) ( )( )PrPr AB.D) Pr(A) + Pr(B) even if A and B are not mutually exclusive.Answer:B3) The cumulative probability distribution shows the probabilityA) that a random variable is less than or equal to a particular value.B) of two or more events occurring at once.C) of all possible events occurring.D) that a random variable takes on a particular value given that another event has happened.Answer:A4) The expected value of a discrete random variable A) is the outcome that is most likely to occur.B) can be found by determining the 50% value in the c.d.f.C) equals the population median.D) is computed as a weighted average of the possible outcome of that random variable, where the weightsare the probabilities of that outcome.Answer:D5) Let Y be a random variable. Then var(Y) equalsA) 2)YE Y 1 ].B)( )YE Y 1 ].C)( ) 2YE Y 1 ].D)( )YE Y 1 ].Answer:C1ScholarStock6) The skewness of the distribution of a random variable Y is defned as follows:A) ( )32YYE Y 1 ]B)( ) 3YE Y 1 ]C) ( )3 33YYE Y 1 ]D) ( ) 33YYE Y 1 ]Answer:D7) The skewness is most likely positive for one of the following distributions:A) The grade distribution at your college or university.B) The U.S. income distribution.C) SAT scores in English.D) The height of 18 year old females in the U.S.Answer:B8) The kurtosis of a distribution is defned as follows:A) ( ) 44YYE Y 1 ]B) ( )4 42YYE Y 1 ]C) var( )skewnessYD) E[(Y -)4)Answer:A9) For a normal distribution, the skewness and kurtosis measures are as follows:A) 1.96 and 4B) 0 and 0C) 0 and 3D) 1 and 2Answer:C2ScholarStock10) The conditional distribution of Y given X = x, Pr(Y = y =x), isA) ( )( )PrPr Y yX x.B)( )1 Pr ,liiX x Y y C) ( )( )Pr ,PrX x Y yY y D) ( )( )Pr ,PrX x Y yX x .Answer:D11) The conditional expectation of Y given X, E(Y , is calculated as follows:A)( )1Prki iiY X x Y y B) EC)( )1Prki iiy Y y X x D)( ) ( )1Prki iiE Y X x X x Answer:C12) Two random variables X and Y are independently distributed if all of the following conditions hold, with the exception ofA) Pr(Y = y= x) = Pr(Y = y).B) knowing the value of one of the variables provides no information about the other.C) if the conditional distribution of Y given X equals the marginal distribution of Y.D) E(Y) = E[E(Y )].Answer:D13) The correlation between X and Y A) cannot be negative since variances are always positive.B) is the covariance squared.C) can be calculated by dividing the covariance between X and Y by the product of the two standard deviations.D) is given by corr(X, Y) = ( )( ) ( )cov ,var varX YX Y.Answer:C3ScholarStock14) Two variables are uncorrelated in all of the cases below, with the exception ofA) being independent.B) having a zero covariance.C) 2 2X YXY D) E(Y) = 0.Answer:C15) var(aX + bY) =A) B) C) D) Answer:B16) To standardize a variable youA) subtract its mean and divide by its standard deviation.B) integrate the area below two points under the normal distribution.C) add and subtract 1.96 times the standard deviation to the variable.D) divide it by its standard deviation, as long as its mean is 1.Answer:A17) Assume that Y is normally distributed N(, 2). Moving from the mean () 1.96 standard deviations tothe left and 1.96 standard deviations to the right, then the area under the normal p.d.f. isA) 0.67B) 0.05C) 0.95D) 0.33Answer:C18) Assume that Y is normally distributed N(, 2). To fnd Pr(c1 Y c2), where c1 < c2 and di = ic ,you need to calculate Pr(d1 Z d2) =A) (d2) - (d1)B) (1.96) - (1.96)C) (d2) - (1 - (d1))D) 1 - ((d2) - (d1))Answer:A4ScholarStock19) If variables with a multivariate normal distribution have covariances that equal zero, thenA) the correlation will most often be zero, but does not have to be.B) the variables are independent.C) you should use the 2 distribution to calculate probabilities.D) the marginal distribution of each of the variables is no longer normal.Answer:B20) The Student t distribution isA) the distribution of the sum of m squared independent standard normal random variables.B) the distribution of a random variable with a chi-squared distribution with m degrees of freedom, divided by m.C) always well approximated by the standard normal distribution.D) the distribution of the ratio of a standard normal random variable, divided by the square root of an independently distributed chi-squared random variable with m degrees of freedom divided by m.Answer:D21) When there are degrees of freedom, the t distributionA) can no longer be calculated.B) equals the standard normal distribution.C) has a bell shape similar to that of the normal distribution, but with "fatter" tails.D) equals the 2Xdistribution.Answer:B22) The sample average is a random variable andA) is a single number and as a result cannot have a distribution.B) has a probability distribution called its sampling distribution.C) has a probability distribution called the standard normal distribution.D) has a probability distribution that is the same as for the Y1,..., Yn i.i.d. variables.Answer:B23) To infer the political tendencies of the students at your college/university, you sample 150 of them. Only one of the following is a simple random sample: YouA) make sure that the proportion of minorities are the same in your sample as in the entire student body.B) call every fftieth person in the student directory at 9 a.m.If the person does not answer the phone, you pick the next name listed, and so on.C) go to the main dining hall on campus and interview students randomly there.D) have your statistical package generate 150 random numbers in the range from 1 to the total number of students in your academic institution, and then choose the corresponding names in the student telephonedirectory.Answer:D5ScholarStock24) The variance of 2, YY , is given by the following formula:A) 2Y .B) Yn.C) 2Yn.D) 2Yn.Answer:C25) The mean of the sample average Y , ( )E Y, isA) 1Yn .B) Y.C) Yn.D) YY for n > 30.Answer:B26) In econometrics, we typically do not rely on exact or fnite sample distributions becauseA) we have approximately an infnite number of observations (think of re-sampling).B) variables typically are normally distributed.C) the covariances of Yi, Yj are typically not zero.D) asymptotic distributions can be counted on to provide good approximations to the exact sampling distribution (given the number of observations available in most cases).Answer:D27) Consistency for the sample average Ycan be defned as follows, with the exception ofA) Yconverges in probability to Y.B) Yhas the smallest variance of all estimators.C) pYY . D) the probability of Ybeing in the range Y c becomes arbitrarily close to one as n increases for any constant c > 0.Answer:B6ScholarStock28) The central limit theorem states that A) the sampling distribution of YYY is approximately normal.B) pYY .C) the probability that Yis in the range Y c becomes arbitrarily close to one as n increases for any constant c > 0.D) the t distribution converges to the F distribution for approximately n > 30.Answer:A29) The central limit theoremA) states conditions under which a variable involving the sum of Y1,..., Yn i.i.d. variables becomes the standard normal distribution.B) postulates that the sample mean Yis a consistent estimator of the population mean Y.C) only holds in the presence of the law of large numbers.D) states conditions under which a variable involving the sum of Y1,..., Yn i.i.d. variables becomes the Student t distribution.Answer:A30) The covariance inequality states thatA) B) C) D) Answer:B31)A) B) C) D) Answer:A7ScholarStock32)( )1niiax b+A) n a x + n bB) n(a + b)C) x n b + D) n a x Answer:A33) Assume that you assign the following subjective probabilities for your fnal grade in your econometrics course (the standard GPA scale of 4 = A to 0 = F applies):Grade ProbabilityA 0.20B 0.50C 0.20D 0.08F 0.02The expected value is:A) 3.0B) 3.5C) 2.78D) 3.25Answer:C34) The mean and variance of a Bernoille random variable are given as A) cannot be calculatedB) np and np(1-p)C) p and( ) 1 p p D) p and (1- p)Answer:D35) Consider the following linear transformation of a random variable y = xxx where x is the mean ofx and x is the standard deviation. Then the expected value and the standard deviation of Y are given asA) 0 and 1B) 1 and 1C) Cannot be computed because Y is not a linear function of XD) x and xAnswer:A8ScholarStock2.2 Essays and Longer Questions1) Think of the situation of rolling two dice and let M denote the sum of the number of dots on the two dice. (So M is a number between 1 and 12.) (a)In a table, list all of the possible outcomes for the random variable M together with its probability distribution and cumulative probability distribution. Sketch both distributions.(b) Calculate the expected value and the standard deviation for M.(c) Looking at the sketch of the probability distribution, you notice that it resembles a normal distribution.Should you be able to use the standard normal distribution to calculate probabilities of events? Why or why not?Answer:(a)(b)7.0; 2.42.(c)You cannot use the normal distribution (without continuity correction) to calculate probabilities of events, since the probability of any event equals zero.9ScholarStock2) What is the probability of the following outcomes? (a) Pr(M = 7)(b) Pr(M = 2 or M = 10)(c) Pr(M = 4 or M 4)(d) Pr(M = 6 and M = 9)(e) Pr(M < 8)(f) Pr(M = 6 or M > 10)Answer:(a) 0.167 or 636 = 16;(b) 0.111 or 439 = 19;(c) 1;(d) 0;(e) 0.583; (f) 0.222 or 836 = 29.10ScholarStock3) Probabilities and relative frequencies are related in that the probability of an outcome is the proportion of the time that the outcome occurs in the long run. Hence concepts of joint, marginal, and conditional probability distributions stem from related concepts of frequency distributions. You are interested in investigating the relationship between the age of heads of households and weekly earnings of households. The accompanying data gives the number of occurrences grouped by age and income. You collect data from 1,744 individuals and think of these individuals as a population that you want to describe, rather than a sample from which you want to infer behavior of a larger population. After sorting the data, you generate the accompanying table:Joint Absolute Frequencies of Age and Income, 1,744 HouseholdsAge of head of household X1X2X3X4 X5The median of the income group of $800 and above is $1,050.(a) Calculate the joint relative frequencies and the marginal relative frequencies. Interpret one of each of these. Sketch the cumulative income distribution.(b) Calculate the conditional relative income frequencies for the two age categories 16-under 20, and 45-under 65. Calculate the mean household income for both age categories. (c) If household income and age of head of household were independently distributed, what would you expect these two conditional relative income distributions to look like? Are they similar here?(d) Your textbook has given you a primary defnition of independence that does not involve conditional relative frequency distributions. What is that defnition? Do you think that age and income are independent here, using this defnition?11ScholarStockAnswer:(a)The joint relative frequencies and marginal relative frequencies are given in the accompanying table.5.2 percent of the individuals are between the age of 20 and 24, and make between $200 and under $400. 21.6 percent of the individuals earn between $400 and under $600.Joint Relative and Marginal Frequencies of Age and Income, 1,744 HouseholdsAge of head of householdX1X2X3X4X5Household Income 16-under 20 20-under 25 25-under 45 45-under 65 65 and > TotalY1$0-under $2000.046 0.044 0.075 0.049 0.014 0.227Y2$200-under $4000.007 0.052 0.198 0.080 0.005 0.342Y3$400-under $6000.000 0.011 0.144 0.058 0.003 0.216Y4$600-under $8000.001 0.006 0.063 0.032 0.001 0.102Y5$800 and >0.001 0.001 0.062 0.048 0.001 0.11212ScholarStock(b)The mean household income for the 16-under 20 age category is roughly $144. It is approximately $489 for the 45-under 65 age category.Conditional Relative Frequencies of Income and Age 16-under 20, and 45-under 65, 1,744 HouseholdsAge of head of householdX1X4Household Income 16-under 2045-under 65Y1$0-under $2000.842 0.185Y2$200-under $400 0.137 0.300Y3$400-under $600 0.000 0.217Y4$600-under $800 0.001 0.118Y5$800 and >0.001 0.180(c) They would have to be identical, which they clearly are not.(d) Pr(Y = y, X = x) = Pr(Y = y) Pr(X = x). We can check this by multiplying two marginal probabilities to see if this results in the joint probability. For example, Pr(Y = Y3) = 0.216 and Pr(X = X3) = 0.542, resulting in a product of 0.117, which does not equal the joint probability of 0.144. Given that we are looking at the data as a population, not a sample, we do not have to test how "close" 0.117 is to 0.144.4) Math and verbal SAT scores are each distributed normally with N (500,10000).(a) What fraction of students scores above 750? Above 600? Between 420 and 530? Below 480? Above 530?(b) If the math and verbal scores were independently distributed, which is not the case, then what would be the distribution of the overall SAT score? Find its mean and variance.(c) Next, assume that the correlation coefcient between the math and verbal scores is 0.75. Find the meanand variance of the resulting distribution.(d) Finally, assume that you had chosen 25 students at random who had taken the SAT exam. Derive the distribution for their average math SAT score. What is the probability that this average is above 530? Why is this so much smaller than your answer in (a)?Answer:(a) Pr(Y>750) = 0.0062; Pr(Y>600) = 0.1587; Pr(420 0 (Y = 0)(u ) 0 (Y = 1)Totalp p1 > 0(X = 0)0.156 0.383 0.539p p1 0(X = 1)0.297 0.164 0.461Total0.453 0.547 1.00(a) Compute E(Y) and E(X), and interpret both numbers.(b) Calculate E(Y= 1) and E(Y= 0). If there was independence between cyclical unemployment and acceleration in the infation rate, what would you expect the relationship between the two expected valuesto be? Given that the two means are diferent, is this sufcient to assume that the two variables are independent?(c) What is the probability of infation to increase if there is positive cyclical unemployment? Negative cyclical unemployment?(d) You randomly select one of the 59 quarters when there was positive cyclical unemployment ((u u ) >0). What is the probability there was decelerating infation during that quarter?Answer:(a) E(Y) = 0.547 . 54.7 percent of the quarters saw cyclical unemployment.E(Y) = 0.461 . 46.1 percent of the quarters saw decreasing infation rates.(b) E(Y= 1) = 0.356; E(Y= 0) = 0.711. You would expect the two conditional expectations to be the same. In general, independence in means does not imply statistical independence, although the reverse is true.(c) There is a 34.4 percent probability of infation to increase if there is positive cyclical unemployment. There is a 70 percent probability of infation to increase if there is negative cyclical unemployment.(d) There is a 65.6 percent probability of infation to decelerate when there is positive cyclical unemployment.20ScholarStock10) The accompanying table shows the joint distribution between the change of the unemployment rate in an election year and the share of the candidate of the incumbent party since 1928. You think of this data as a population which you want to describe, rather than a sample from which you want to infer behavior of a larger population. Joint Distribution of Unemployment Rate Change and Incumbent Party's VoteShare in Total Vote Cast for the Two Major-Party Candidates,1928-2000(Incumbent- 50%) > 0 (Y = 0)(Incumbent- 50%) 0(Y = 1)Totalu > 0 (X = 0) 0.053 0.211 0.264u 0 (X = 1) 0.579 0.157 0.736Total0.632 0.368 1.00(a) Compute and interpret E(Y) and E(X).(b) Calculate E(Y= 1) and E(Y= 0). Did you expect these to be very diferent?(c) What is the probability that the unemployment rate decreases in an election year?(d) Conditional on the unemployment rate decreasing, what is the probability that an incumbent will losethe election?(e) What would the joint distribution look like under independence?Answer:(a) E(Y) = 0.368; E(X) = 0.736. The probability of an incumbent to have less than 50% of the share of votes cast for the two major-party candidates is 0.368. The probability of observing falling unemployment rates during the election year is 73.6 percent.(b) E(Y= 1) = 0.213; E(Y= 0) = 0.799. A student who believes that incumbents will attempt to manipulate the economy to win elections will answer afrmatively here.(c) Pr(X = 1) = 0.736.(d) Pr(Y = 1= 1) = 0.213.(e) Joint Distribution of Unemployment Rate Change and Incumbent Party's Vote Share in Total Vote Cast for the Two Major-Party Candidates,1928-2000 under Assumption of Statistical Independence(Incumbent- 50%) > 0(Y = 0)(Incumbent- 50%) > 0(Y = 1)Totalu > 0 (X = 0) 0.167 0.097 0.264u 0 (X = 1) 0.465 0.271 0.736Total0.632 0.368 1.0021ScholarStock11) The table accompanying lists the joint distribution of unemployment in the United States in 2001 by demographic characteristics (race and gender).Joint Distribution of Unemployment by Demographic Characteristics, United States, 2001White(Y = 0)Black and Other(Y = 1)TotalAge 16-19(X = 0)0.13 0.05 0.18Age 20 and above(X = 1)0.60 0.22 0.82Total 0.73 0.27 1.00(a) What is the percentage of unemployed white teenagers?(b) Calculate the conditional distribution for the categories "white" and "black and other."(c) Given your answer in the previous question, how do you reconcile this fact with the probability to be 60% of fnding an unemployed adult white person, and only 22% for the category "black and other."Answer:(a) Pr(Y = 0, X = 0) = 0.13.(b) Conditional Distribution of Unemployment by Demographic Characteristics, United States, 2001White(Y = 0)Black and Other(Y = 1)Age 16-19(X = 0)0.18 0.19Age 20 and above(X = 1)0.82 0.81Total 1.00 1.00(c) The original table showed the joint probability distribution, while the table in (b) presented the conditional probability distribution.22ScholarStock12) From the Stock and Watson (http://www.pearsonhighered.com/stock_watson) website the chapter 8 CPS data set (ch8_cps.xls) into a spreadsheet program such as Excel. For the exercise, use the frst 500 observations only. Using data for average hourly earnings only (ahe), describe the earnings distribution. Use summary statistics, such as the mean, median, variance, and skewness. Produce a frequency distribution ("histogram") using reasonable earnings class sizes.Answer:aheMean 19.79Standard Error0.51Median 16.83Mode 19.23Standard Deviation 11.49Sample Variance 131.98Kurtosis 0.23Skewness 0.96Range 58.44Minimum 2.14Maximum 60.58Sum 9897.45Count 500.0The mean is $19.79. The median ($16.83) is lower than the average, suggesting that the mean is being pulled up by individuals with fairly high average hourly earnings. This is confrmed by the skewness measure, which is positive, and therefore suggests a distribution with a long tail to the right. The varianceis $2131.96, while the standard deviation is $11.49.To generate the frequency distribution in Excel, you frst have to settle on the number of class intervals. Once you have decided on these, then the minimum and maximum in the data suggests the class width. In Excel, you then defne "bins" (the upper limits of the class intervals). Sturges's formula can be used to suggest the number of class intervals (1+3.31log(n) ), which would suggest about 9 intervals here. Instead I settled for 8 intervals with a class width of $8 minimum wages in California are currently $8 and approximately the same in other U.S. states.23ScholarStockThe table produces the absolute frequencies, and relative frequencies can be calculated in a straightforward way.bins Frequency rel. freq.8 50 0.116 187 0.37424 115 0.2332 68 0.13640 38 0.07648 33 0.06656 8 0.01666 1 0.002More 0Substitution of the relative frequencies into the histogram table then produces the following graph (after eliminating the gaps between the bars).24ScholarStock2.3 Mathematical and Graphical Problems1) Think of an example involving fve possible quantitative outcomes of a discrete random variable and attach a probability to each one of these outcomes. Display the outcomes, probability distribution, and cumulative probability distribution in a table. Sketch both the probability distribution and the cumulativeprobability distribution.Answer:Answers will vary by student. The generated table should be similar to Table 2.1 in the text, and fgures should resemble Figures 2.1 and 2.2 in the text.2) The height of male students at your college/university is normally distributed with a mean of 70 inchesand a standard deviation of 3.5 inches. If you had a list of telephone numbers for male students for the purpose of conducting a survey, what would be the probability of randomly calling one of these students whose height is(a) taller than 6'0"?(b) between 5'3" and 6'5"?(c) shorter than 5'7", the mean height of female students?(d) shorter than 5'0"?(e) taller than Shaquille O'Neal, the center of the Boston Celtics, who is 7'1" tall? Compare this to the probability of a woman being pregnant for 10 months (300 days), where days of pregnancy is normally distributed with a mean of 266 days and a standard deviation of 16 days. Answer:(a) Pr(Z > 0.5714) = 0.2839;(b) Pr( 2 < Z < 2) = 0.9545 or approximately 0.95; (c) Pr(Z < -0.8571) = 0.1957; (d) Pr(Z < -2.8571) = 0.0021; (e) Pr(Z > 4.2857) = 0.000009 (the text does not show values above 2.99 standard deviations, Pr(Z>2.99 = 0.0014) and Pr(Z > 2.1250) = 0.0168.25ScholarStock3) Calculate the following probabilities using the standard normal distribution. Sketch the probability distribution in each case, shading in the area of the calculated probability. (a) Pr(Z < 0.0)(b) Pr(Z 1.0)(c) Pr(Z > 1.96)(d) Pr(Z < 2.0)(e) Pr(Z > 1.645)(f) Pr(Z > 1.645)(g) Pr(1.96 < Z < 1.96)(h.) Pr(Z < 2.576 or Z > 2.576)(i.) Pr(Z > z) = 0.10; fnd z.(j.) Pr(Z < z or Z > z) = 0.05; fnd z.Answer:(a) 0.5000; (b) 0.8413; (c) 0.0250; (d) 0.0228; (e) 0.0500; (f) 0.9500; (g) 0.0500; (h) 0.0100; (i) 1.2816; (j) 1.96.26ScholarStock4) Using the fact that the standardized variable Z is a linear transformation of the normally distributed random variable Y, derive the expected value and variance of Z.Answer:Z = 1Y YY Y YY +Y = a + bY, with a = - YY and b = 1Y.Given (2.29) and (2.30) in the text, E(Z) = - YY + 1YY = 0, and 2211Z ZZ .5) Show in a scatterplot what the relationship between two variables X and Y would look like if there was(a) a strong negative correlation.(b) a strong positive correlation.(c) no correlation.Answer: (a)27ScholarStock(b)(c)6) What would the correlation coefcient be if all observations for the two variables were on a curve described by Y = X2?Answer:The correlation coefcient would be zero in this case, since the relationship is non-linear.28ScholarStock7) Find the following probabilities:(a) Y is distributed 24X. Find Pr(Y > 9.49).(b) Y is distributed t. Find Pr(Y > 0.5).(c) Y is distributed F4, . Find Pr(Y < 3.32).(d) Y is distributed N(500, 10000). Find Pr(Y > 696 or Y < 304).Answer:(a) 0.05.(b) 0.6915.(c) 0.99.(d) 0.05.8) In considering the purchase of a certain stock, you attach the following probabilities to possible changes in the stock price over the next year.Stock Price Change During Next Twelve Months (%)Probability+15 0.2+5 0.300.450.05150.05What is the expected value, the variance, and the standard deviation? Which is the most likely outcome? Sketch the cumulative distribution function.Answer:E(Y) = 3.5; 2Y = 8.49; Y = 2.91; most likely: 0.29ScholarStock9) You consider visiting Montreal during the break between terms in January. You go to the relevant Web site of the ofcial tourist ofce to fgure out the type of clothes you should take on the trip. The site lists that the average high during January is 7 C, with a standard deviation of 4 C. Unfortunately you are more familiar with Fahrenheit than with Celsius, but fnd that the two are related by the following linear function: C=59(F 32).Find the mean and standard deviation for the January temperature in Montreal in Fahrenheit.Answer:Using equations (2.29) and (2.30) from the textbook, the result is 19.4 and 7.2.10) Two random variables are independently distributed if their joint distribution is the product of their marginal distributions. It is intuitively easier to understand that two random variables are independently distributed if all conditional distributions of Y given X are equal. Derive one of the two conditions from the other.Answer:If all conditional distributions of Y given X are equal, then( ) Pr 1 Pr( 2) Pr( 1) Y y X Y y X Y y X KBut if all conditional distributions are equal, then they must also equal the marginal distribution, i.e.,( ) Pr Pr( ) Y y X x Y y Given the defnition of the conditional distribution of Y given X = x, you then getPr(Y = y= x) = Pr( , )Pr( )Y y X xX x = Pr(Y = y),which gives you the condition Pr(Y = y, X = x) = Pr(Y = y) Pr(X = x).11) There are frequently situations where you have information on the conditional distribution of Y given X, but are interested in the conditional distribution of X given Y. Recalling Pr(Y = y= x) =Pr( , )Pr( )X x Y yX x , derive a relationship between Pr(X = x= y) and Pr(Y = y= x). This is called Bayes' theorem.Answer:Given Pr(Y = y= x) = Pr( )Pr( )X x Y yX x ,Pr(Y = y= x) Pr(X = x) = Pr(X = x, Y = y);similarly Pr(X = x= y) = Pr( )Pr( )X x Y yX x andPr(X = x= y) Pr(Y = y) = Pr(X = x, Y = y). Equating the two and solving for Pr(X = x= y) then results inPr(X = x= y) = Pr( ) Pr( )Pr( )Y y X x X xY y .30ScholarStock12) You are at a college of roughly 1,000 students and obtain data from the entire freshman class (250 students) on height and weight during orientation. You consider this to be a population that you want to describe, rather than a sample from which you want to infer general relationships in a larger population. Weight (Y) is measured in pounds and height (X) is measured in inches. You calculate the following sums:= 94,228.8,= 1,248.9,= 7,625.9(small letters refer to deviations from means as in= ). (a) Given your general knowledge about human height and weight of a given age, what can you say aboutthe shape of the two distributions? (b) What is the correlation coefcient between height and weight here?Answer:(a) Both distributions are bound to be normal.(b) 0.703.13) Use the defnition for the conditional distribution of Y given X = x and the marginal distribution of X to derive the formula for Pr(X = x, Y = y). This is called the multiplication rule. Use it to derive the probability for drawing two aces randomly from a deck of cards (no joker), where you do not replace the card after the frst draw. Next, generalizing the multiplication rule and assuming independence, fnd the probability of having four girls in a family with four children.Answer:452 351 = 0.0045; 0.0625 or 41 12 16 _ _ , ,.14) The systolic blood pressure of females in their 20s is normally distributed with a mean of 120 with a standard deviation of 9. What is the probability of fnding a female with a blood pressure of less than 100? More than 135? Between 105 and 123? You visit the women's soccer team on campus, and fnd that the average blood pressure of the 25 members is 114. Is it likely that this group of women came from the same population?Answer:Pr(Y135) = 0.0478; Pr(105 0.25). 33ScholarStock19) The accompanying table lists the outcomes and the cumulative probability distribution for a student renting videos during the week while on campus.Video Rentals per Week during SemesterOutcome (number of weeklyvideo rentals)0 1 2 3 4 5 6Probability distribution 0.05 0.55 0.25 0.05 0.07 0.02 0.01Sketch the probability distribution. Next, calculate the cumulative probability distribution for the above table. What is the probability of the student renting between 2 and 4 a week? Of less than 3 a week?Answer:The cumulative probability distribution is given below. The probability of renting between two and four videos a week is 0.37. The probability of renting less than three a week is 0.85.Outcome (number of weekly video rentals)0 1 2 3 4 5 6Cumulative probability distribution0.05 0.60 0.85 0.90 0.97 0.99 1.0020) The textbook mentioned that the mean of Y, E(Y) is called the frst moment of Y, and that the expectedvalue of the square of Y, E(Y2) is called the second moment of Y, and so on. These are also referred to as moments about the origin. A related concept is moments about the mean, which are defned as E[(Y Y)r]. What do you call the second moment about the mean? What do you think the third moment, referred to as "skewness," measures? Do you believe that it would be positive or negative for an earnings distribution? What measure of the third moment around the mean do you get for a normal distribution? Answer:The second moment about the mean is the variance. Skewness measures the departure from symmetry. For the typical earnings distribution, it will be positive. For the normal distribution, it will be zero.34ScholarStock21) Explain why the two probabilities are identical for the standard normal distribution: Pr(1.96 X 1.96) and Pr(1.96 < X < 1.96).Answer:For a continuous distribution, the probability of a point is zero.22) SAT scores in Mathematics are normally distributed with a mean of 500 and a standard deviation of 100. The formula for the normal distribution is. Use the scatter plot option in a standard spreadsheet program, such as Excel, to plot the Mathematics SAT distribution using this formula. Start by entering 300 as the frst SAT score in the frst column (the lowest score you can get in the mathematics section as long as you fll in your name correctly), and then increment the scores by 10 until you reach 800. In the second column, use the formula for the normal distribution and calculate f(Y). Then use the scatter plot option, where you eventually remove markers and substitute these with the solid line option.Answer:23) Use a standard spreadsheet program, such as Excel, to fnd the following probabilities from various distributions analyzed in the current chapter:a. If Y is distributed N (1,4), fnd Pr(Y 3)b. If Y is distributed N (3,9), fnd Pr(Y > 0)c. If Y is distributed N (50,25), fnd Pr(40 Y 52)d. If Y is distributed N (5,2), fnd Pr(6 Y 8)Answer:The answers here are given together with the relevant Excel commands.a.=NORMDIST(3,1,2,TRUE) = 0.8413b.=1-NORMDIST(0,3,3,TRUE) = 0.8413c.=NORMDIST(52,50,5,TRUE)-NORMDIST(40,50,5,TRUE) = 0.6326d.=NORMDIST(8,5,SQRT(2),TRUE)-NORMDIST(6,5,SQRT(2),TRUE) = 0.222935ScholarStock24) Looking at a large CPS data set with over 60,000 observations for the United States and the year 2004, you fnd that the average number of years of education is approximately 13.6. However, a surprising large number of individuals (approximately 800) have quite a low value for this variable, namely 6 years or less.You decide to drop these observations, since none of your relatives or friends have that few years of education. In addition, you are concerned that if these individuals cannot report the years of education correctly, then the observations on other variables, such as average hourly earnings, can also not be trusted. As a matter of fact you have found several of these to be below minimum wages in your state. Discuss if dropping the observations is reasonable.Answer:While it is always a good idea to check the data carefully before conducting a quantitative analysis, you should never drop data before carefully thinking about the problem at hand. While it is not plausible to fnd many individuals in the U.S. who were raised here with that few years of education, there will be immigrants in the survey. Average years of education can be quite low in other countries. For example, Brazil's average years of schooling is less than 6 years. The point of the exercise is to think hard whether or not observations are outliers generated by faulty data entry or if there is a reason for observing values which may appear strange at frst.25) Use a standard spreadsheet program, such as Excel, to fnd the following probabilities from various distributions analyzed in the current chapter:a.If Y is distributed 24X, fnd Pr(Y 7.78)b.If Y is distributed 210X, fnd Pr(Y > 18.31)c. If Y is distributed F10,, fnd Pr(Y > 1.83)d.If Y is distributed t15, fnd Pr(Y > 1.75)e. If Y is distributed t90, fnd Pr(-1.99 Y 1.99)f. If Y is distributed N(0,1), fnd Pr(-1.99 Y 1.99)g.If Y is distributed F10,4, fnd Pr(Y > 4.12)h.If Y is distributed F7,120, fnd Pr(Y > 2.79)Answer:The answers here are given together with the relevant Excel commands.a.=1-CHIDIST(7.78,4) = 0.90b.=CHIDIST(18.31,10) = 0.05c.=FDIST(1.83,10,1000000) = 0.05d.=TDIST(1.75,15,1) = 0.05e.=1-TDIST(1.99,90,2) = 0.95f. =NORMDIST(1.99,0,1,1)-NORMDIST(-1.99,0,1,1) = 0.953g.=FDIST(4.12,7,4) = 0.10h.=FDIST(2.79,7,120) = 0.0136ScholarStock